Complex Series Solution
For what value of $ n $ is $ i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i $?
Note: $ i^2 = -1 $
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Solution
The real numbers in the sum follow the pattern
\[-2 + 4 - 6 + 8 - 10 + 12 + \dotsb.\]Placing the numbers in pairs, we get
\[(-2 + 4) + (-6 + 8) + (-10 + 12) + \dotsb = 2 + 2 + 2 + \dotsb.\]Thus, we want 24 pairs. The second number in the $ n $th pair is $ 4n, $ which means the last real term is $ 96 = 96i^{96} $. Thus, $ n $ is either 96 or 97.
Up to $ 96i^{96}, $ the sum of the imaginary numbers is
\[i - 3i + 5i - 7i + 9i - 11i + \dots + 93i - 95i.\]Similarly, if we pair these numbers, we get
\[(i - 3i) + (5i - 7i) + (9i - 11i) + \dots + (93i - 95i) = -2i - 2i - 2i - \dots - 2i = -48i.\]Adding $ 97i^{97} = 97i, $ we get $ 97i - 48i = 49i $. Hence, $ n = \boxed{97} $.
Freeflows